Asked by Kyle

A charge Q = +4.5 uC is located at the origin and is fixed so that it cannot move. A small object with mass m = 2.0 g and charge q = +8.0 uC is released from rest at point A on the x-axis at x = 2.0 cm. The small object moves without friction along the x-axis under the influence of the charge Q only (gravity can be neglected) and eventually reaches point B on the x-axis at x = 10.0 cm. (a) Calculate the potential difference between point B and point A (change in VAB = VB-VA). (b) Calculate the speed of the small object when it reaches point B
Answered by Elena
Δφ=φ₁-φ₂=kQ/r₁-kQ/r₂=
=kQ(r₂-r₁)/r₁•r₂=
=9•10⁹•4.5•10⁻⁶•8•10⁻²/20•10⁻⁴=1.62•10⁶ V
mv²/2= q•Δφ,
v= sqrt(2•q•Δφ/m)=…
Answered by Meagan
Elena,
Im a little stuck on the first equation you have given. In the denominator, we are supposed to multiply r1 by r2. How did you get 20 x 10? Shouldn't it be 0.1m x 0.02m?
Answered by Meagan
Nevermind, that'd be my error. Read it wrong
Answered by Kyle
Shouldn't the answer be 1.62*10^4V? Unless i'm putting the numbers in wrong. Please help
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